Math Analysis Semester 2 Study Guide

Math Analysis Semester 2 Study Guide Roots/ Polynomials: If f(x) is a polynomial of degree n where n>0, then f has at least one zero in the complex # system. Know Rational Root Theorem: (factors of last co-efficient/factors of 1st co-efficient) Know Polynomial Division: Ex.

fx= x3+11x2+39x+29

Use Rational Root Theorem… to get: ±29, ±1 as factors -1

1 1

11 -1 10

39 29 -10 -29 29 0

(x2+10x+29)(x+1) Then use the quadratic formula

Rational Functions and Asymptotes: Vertical Asymptotes: where denominator = 0 (only after factoring and canceling) Holes (deleted pts): are what cancels out Horizontal Asymptotes: the # you approach as |x| dominant term then cancel) Steps:

∞ (only look @ the

1.factor numerator + denominator 2. Write restrictions (what makes denominator = 0) 3. Cancel (the holes) 4. Denominator = 0 after canceling are the vertical asymptotes.

Ex. fx= x2+2x2x2-x = xx+2x2x-1 x≠0,12 fx= x+22x-1 hole:x=0 Vertical Asymptote:x=12 Horixontal Asymptote:y=12 ***BE ABLE TO GRAPH BY HAND AND STATE THE DOMAIN***

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

Slant Asymptotes: only occur when power in numerator is ONE larger than power in denominator To Find: use Long Division fx= x3+x2-2x-2x2+ 3 Y= x+1 +-5X-4x2+3 x2+3 x3+x2-2x-1 x3 3x x2-5x-1 x2 3 -5x-4 Slant:

Sequences A sequence is a function whose domain is the set of positive integers Ex. An=3n-2 sequence is 1,4,7,10… Factorials review: n! = 1·2·3...n 0!=1 5!3!=5×4=20 Summation notation: i=1nai=a1+a2…+an Example: i=372i-1=5+7+9+11+13=45 Arithmetic Sequences A1,A2,A3…,An A2 – A1 =d A3 – A2 =d you get the idea d is the common difference 1, 4, 7, 10 d=3 an=a1+(n-1)d Sum of finite arithmetic sequence: Sn= n2(a1+an)=n(a1+an2) Geometric Sequences: A2/ A1 = A3/ A2=r An=a1rn-1 Sum of a Finite Geometric Sequence Sn=1-rn1-r= a1-arn1-r Sum of an infinite geometric sequence S∞=a11-r Where |r|<1

Study Guide of Examples for Sequences and Functions: Linear: lim 3x+1 = 3(2)+1 = 7 © 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

x 2

Proof:

7-ϵ<3x+1<7+ϵ 6-ϵ<3x<6+ϵ 2-ϵ3<x<2+ϵ3 δ=ϵ3

Choose δ = E/3 since all steps are reversible whenever x E <2-δ, 2+δ> and x 2 then 3x+1 E <7-E, 7+E> and the limit is 7

Quadratics: Lim x2+5 = (2)2 +5 = 9 x2 |( x2 +5)-(9)|<E | x2 – 4|<E |x+2| |x-2|<E let ≤ 1 |x-2| <1 -1<x-2<1 3<x+2<5 |x+2|<5 |x+2| |x-2|<E |x-2|(5)<E |x-2|<E/5 δ<E/5 Rationals: Lim (x+4)/(x-2) = (3+4)/(3-2) = 7 x 3 |( x+4)/(x-2) – 7|<E |[x+4-7(x-2)]/(x-2)|<E |(x+4-7x+14)/(x-2)|<E |(18-6x)/(x-2)|<E |-6| |x-3| |1/(x-2)| <E |x-3| |1/(x-2)|<E/6 let ≤ ½ |x-3|<1/2 -1/2 < x-3 < 1/2 ½ < x-2 < 3/2 2 > 1/(x-2) > 2/3 |1/(x-2)|<2 |x-3| |1/(x-2)|<E/6 |x-3| (2) <E/6 |x-3|<E/12

Proof: Let ≤1 then |x-2|<1 and |x+2|<5. If we also let |x-2|<E/5 then the product: |x+2| |x-2|<E |x2 – 4|<E |( x2 +5)-(9)|<E so choose =1 or E/5 whichever is smaller than whenever x E <2-δ, 2+ > and x 2 then x2+5 E <9-E, 9+E> and the limit is 9

Proof: Let ≤ ½ then |x-3|<1/2 and |1/(x-2)|<2. If we also let |x-3| <E/12 then the product: |x-3| |1/(x-2)|<E/6 |-6| |x-3| |1/(x-2)| <E |(18-6x)/(x-2)|<E |(x+4-7x+14)/(x-2)|<E |[x+4-7(x-2)]/(x-2)|<E |( x+4)/(x-2) – 7|<E so choose δ=1/2 or E/12 whichever is smaller then whenever x E <3- , 3+δ> and x 3 then (x+4)/(x-2) E <7-E, 7+E> and the limit is 7

X ∞ Lim (x+1)/(x-1) = 1 © 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

x ∞ Lim (x+1)/(x-1) = 1 x+∞ ***notice the positive sign 1-E< (x+1)/(x-1) < 1+E (x-1)(1-E)<x+1 1-xE-1+E<x+1 -2/E + 1<x true x+1< (1+E)(x-1) x+1<x-1+Ex-E 2/E + E/E < E/E –x 2/E +1 <x Lim (x+1)/(x-1) = 1 x-∞ ***notice the negative sign 1-E< (x+1)/(x-1) < 1+E (1-E)(x-1)>x+1 ** here x is a negative so you flip the sign** x-1-Ex+E>x+1 -1-Ex+E>1 -Ex>2-E x<-2/E +1 x+1>(1+E)(x-1) x+1>x-1+Ex-E 1>1+Ex-E 2+E>Ex x<2/E +1 true since x-∞ so x is a negative Proof: Choose |x0| = |2/E +1| **choose the value farther from zero if there are 2 different answers** then for all | x|>|x0| (x+1)/(x-1) E <1-E, 1+E> and the limit is 1.

5.3 Properties of Logs y = loga x if x = ay remember e is just the natural log, and can be used like any other number so

loge x = ln x

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

loga x =

change of base

logb x logx ln x = = logb a loga ln a

loga (uv) = loga u + loga v loga

u = loga u − loga v v

loga u n = n loga u

ln(uv) = ln(u) + ln(v) u ln( ) = ln(u) − ln(v) v

lnu n = n lnu and the log of 1 is 0 ***remember to use these rules to solve logarithmic equations

3.1 Exponential Functions P is the initial investment and r is the interest rate (write % as

A = Pe rt decimal) this formula is used for continuous compounding which means that the investment is compounded an infinite amount of times per year (t). A is the ending amount For n compoundings per year:

r A = P(1+ ) nt n

t is the number of years and n is the number of times per year the

amount P is compounded. R is the rate of interest (write 3% as .03) Annuities: some payment over a length of time PV= Present value © 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan

© 2009 Math Analysis Semester 2: Maya Balakrishnan and Tara Balakrishnan